Information

EE3376

Topics to Cover…

l  Binary Digital System l  Data Types l  Conversions l  Binary Arithmetic l  Overflow l  Logical Operations l  Fixed Point l  Floating Point l  ASCII Characters

Adapted from notes from BYU ECE124

2

What are Decimal Numbers?

l “Decimal” means that we have ten digits to use in our representation

–  the symbols 0 through 9 l What is 3,546?

–  3 thousands + 5 hundreds + 4 tens + 6 ones. –  3,54610 = 3×103 + 5×102 + 4×101 + 6×100

l How about negative numbers? –  Use two more symbols to distinguish positive and

negative, namely, + and -.

Adapted from notes from BYU ECE124 Adapted from notes from BYU ECE124

3

What are Binary Numbers?

l “Binary” means that we have two digits to use in our representation

–  the symbols 0 and 1 l What is 1011?

–  1 eights + 0 fours + 1 twos + 1 ones –  10112 = 1×23 + 0×22 + 1×21 + 1×20

l How about negative numbers? –  We don’t want to add additional symbols –  So…

Adapted from notes from BYU ECE124

4

Electronic Representation of a Bit

l  Relies only on approximate physical values. –  A logical ‘1’ is a relatively high voltage (2.4V - 5V). –  A logical ‘0’ is a relatively low voltage (0V - 1V).

n  Analog processing relies on exact values which are affected by temperature, age, etc. n  Analog values are never quite the same. n  Each time you play a vinyl album, it will sound a bit different. n  CDs sound the same no matter how many times you play them.

Adapted from notes from BYU ECE124

5

The Power of the Bit…

l  Bits rely on approximate physical values that are not affected by age, temperature, etc. –  Music that never degrades. –  Pictures that never get dusty or scratched.

l  By using groups of bits, we can achieve high precision.

–  8 bits => each pattern represents 1/256. –  16 bits => each pattern represents 1/65,536 –  32 bits => each pattern represents 1/4,294,967,296 –  64 bits => each pattern represents 1/18,446,744,073,709,550,000

l  Disadvantage: bits only represent discrete values l  Digital = Discrete

Adapted from notes from BYU ECE124 6

Binary Nomenclature l  Binary Digit: 0 or 1 l  Bit (short for binary digit): A single binary digit l  LSB (least significant bit): The rightmost bit l  MSB (most significant bit): The leftmost bit l  Data sizes

–  1 Nibble = 4 bits –  1 Byte = 2 nibbles = 8 bits –  1 Kilobyte (KB) = 1024 bytes –  1 Megabyte (MB) = 1024 kilobytes = 1,048,576 bytes –  1 Gigabyte (GB) = 1024 megabytes = 1,073,741,824

bytes Adapted from notes from BYU ECE124

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What Kinds of Data? l  All kinds…

–  Numbers – signed, unsigned, integers, floating point, complex, rational, irrational, …

–  Text – characters, strings, … –  Images – pixels, colors, shapes, … –  Sound – pitch, amplitude, … –  Logical – true / false, open / closed, on / off, … –  Instructions – programs, … –  …

l  Data type: –  representation and operations within the computer

l  We’ll start with numbers… Adapted from notes from BYU ECE124

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Some Important Data Types

l  Unsigned integers –  only non-negative numbers –  0, 1, 2, 3, 4, …

l  Signed integers –  negative, zero, positive numbers –  …, -3, -2, -1, 0, 1, 2, 3, …

l  Floating point numbers –  numbers with decimal point –  PI = 3.14159 x 100

l  Characters –  8-bit, unsigned integers –  ‘0’, ‘1’, ‘2’, … , ‘a’, ‘b’, ‘c’, … , ‘A’, ‘B’, ‘C’, … , ‘@’, ‘#’,

Adapted from notes from BYU ECE124

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Unsigned Integers

329 102 101 100

101 22 21 20

3x100 + 2x10 + 9x1 = 329 1x4 + 0x2 + 1x1 = 5

most significant

least significant

l  What do these unsigned binary numbers represent? 0000 0110

1111 1010

0001 1000

0111 1100

1011 1001

n  Weighted positional notation n  “3” is worth 300, because of its position, while “9” is only worth 9

Adapted from notes from BYU ECE124

10

Unsigned Integers (continued…)

22 21 20 0 0 0 0 0 0 1 1 0 1 0 2 0 1 1 3 1 0 0 4 1 0 1 5 1 1 0 6 1 1 1 7

Data Types

Adapted from notes from BYU ECE124

11

Unsigned Binary Arithmetic

l  Base 2 addition – just like base 10! –  add from right to left, propagating carry

10010 10010 1111 + 1001 + 1011 + 1

11011 11101 10000

10111 + 111

carry

0 1 1 1 1 Adapted from notes from BYU ECE124

12

Signed Integers

l  With n bits, we have 2n distinct values. –  assign about half to positive integers (1 through 2n-1)

and about half to negative (- 2n-1 through -1) –  that leaves two values: one for 0, and one extra

l  Positive integers –  just like unsigned – zero in most significant (MS) bit

00101 = 5 l  Negative integers

–  sign-magnitude – set MS bit to show negative 10101 = -5

–  one’s complement – flip every bit to represent negative 11010 = -5

–  MS bit indicates sign: 0=positive, 1=negative Adapted from notes from BYU ECE124

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2’s Complement

l  Problems with sign-magnitude and 1’s complement –  two representations of zero (+0 and –0) –  arithmetic circuits are complex

l  How to add two sign-magnitude numbers? e.g., try 2 + (-3)

l  How to add to one’s complement numbers? e.g., try 4 + (-3)

l  Two’s complement representation developed to make circuits easy for arithmetic.

Adapted from notes from BYU ECE124

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2’s Complement (continued…)

l  Simplifies logic circuit construction because –  addition and subtraction are always done using the

same circuitry. –  there is no need to check signs and convert. –  operations are done same way as in decimal

l  right to left l  with carries and borrows

l  Bottom line: simpler hardware units!

Adapted from notes from BYU ECE124

15

2’s Complement (continued…)

l  If number is positive or zero, –  normal binary representation

l  If number is negative, –  start with positive number –  flip every bit (i.e., take the one’s complement) –  then add one

00101 (5) 01001 (9)

11010 (1’s comp) (1’s comp)

+ 1 + 1

11011 (-5) (-9)

10110

10111 Adapted from notes from BYU ECE124

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2’s Complement (continued…)

l  Positional number representation with a twist –  the most significant (left-most) digit has a negative

weight

–  n-bits represent numbers in the range -2n-1 … 2n-1 - 1 l  What are these?

0110 = 22 + 21 = 6

1110 = -23 + 22 + 21 = -2

0000 0110 1111 1010 0001

1000 0111 1100 1011 1001

0121 2222 −−− nn

Adapted from notes from BYU ECE124

17

Number Decimal Value Negated Binary Value011001110000111101001000

2’s Complement Negation

To negate a number, invert all the bits and add 1

6 1010

7 1001

0 0000

-1 0001

4 1100

-8 1000 (??)

Adapted from notes from BYU ECE124

18

Two’s complement – simple conversion

0001102 = 610 0001102 Find first “1” from the right

0000102 invert each digit to the left

0010102 0110102 1110102 = -610 until you reach the left end

To obtain negative 6, start with positive magnitude and negate

Adapted from notes from BYU ECE124

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l  Continually divide the number by 2 and track the remainders.

1 × 25 + 0 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 1 × 20 32 + 0 + 8 + 0 + 2 + 1 = 43

43 2

2

2

2

2

2

Decimal to Binary Conversion

n  For negative numbers, do above for positive number and negate result

5 R 0

0

2 R 1

1 21 R 1

1

10 R 1

1

1 R 0

0

0 R 1

1

Adapted from notes from BYU ECE124

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Number Binary Value56

12335

-351007

Decimal to Binary Conversion

0101

0110

01111011

00100011

11011101

01111101111

Adapted from notes from BYU ECE124

21

Sign-Extension in 2’s Complement

l  You can make a number wider by simply replicating its leftmost bit as desired.

0110 = 000000000000000110 = 1111 = 11111111111111111 = 1 =

6

6

-1

-1 -1

n  What do these represent?

Adapted from notes from BYU ECE124

22

Hexadecimal Notation

l  Binary is hard to read and write by hand l  Hexadecimal is a common alternative

–  16 digits are 0123456789ABCDEF

0100 0111 1000 1111 = 0x478F 1101 1110 1010 1101 = 0xDEAD 1011 1110 1110 1111 = 0xBEEF 1010 0101 1010 0101 = 0xA5A5

Binary Hex

0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F

0x is a common prefix for writing numbers which mean hexadecimal

1.  Separate binary code into groups of 4 bits (starting from the right)

2.  Translate each group into a single hex digit

Adapted from notes from BYU ECE124

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Why Hexadecimal? l  One Hexadecimal digit

represents four binary digits l  Hexadecimal is more human

readable l  Can quickly convert large

binary numbers to hexadecimal and back by inspection

l  Nomenclature –  Motorola = $64 –  C = 0x64 –  Intel = 64H –  TI = 64h

Decimal Hexadecimal Binary 0 0 0000 1 1 0001 2 2 0010 3 3 0011 4 4 0100 5 5 0101 6 6 0110 7 7 0111 8 8 1000 9 9 1001

10 a 1010 11 b 1011 12 c 1100 13 d 1101 14 e 1110 15 f 1111 Adapted from notes from BYU ECE124

24

Binary to Hexadecimal – Easy!

1011

B

0001

1

1100

C

0101

5

So $b1c5 equals %1011000111000101.

25

Decimal to Hex Examples

12decimal = 1100 = 0xc

21decimal = 0001 0101 = 0x15

55decimal = 0011 0111 = 0x37

256decimal = 0001 0000 0000 = 0x100

47decimal = 0010 1111 = 0x2f

3decimal = 0011 = 0x3

127decimal = 0111 1111 = 0x7f

1029decimal = 0100 0000 0101 = 0x405

Adapted from notes from BYU ECE124

26

Word Sizes

l  In the preceding slides, every bit pattern was a different length (15 was represented as 01111).

l  Every real computer has a base word size –  our machine (MPS430) is 16-bits

l  Memory fetches are word-by-word –  even if you only want 8 bits (a byte)

l  Instructions are packed into words l  Numeric representations are word-sized

–  15 is represented as 0000000000001111

Conversions

Adapted from notes from BYU ECE124

27

Number Wheel for 3 bit hardware

000

001

010

011

100

101

110

111

0 unsigned or signed

1 unsigned or signed

2 unsigned or signed

3 unsigned or signed

4 unsigned / -4 signed 5 unsigned / -3 signed

6 unsigned / -2 signed

7 unsigned / -1 signed

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Number Wheel – Unsigned Overflow

000

001

010

011

100

101

110

111

0 unsigned or signed

1 unsigned or signed

2 unsigned or signed

3 unsigned or signed

4 unsigned / -4 signed 5 unsigned / -3 signed

6 unsigned / -2 signed

7 unsigned / -1 signed

6 + 3 = 9 (too large for 3 bits)

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Number Wheel – Signed Overflow

000

001

010

011

100

101

110

111

0 unsigned or signed

1 unsigned or signed

2 unsigned or signed

3 unsigned or signed

4 unsigned / -4 signed 5 unsigned / -3 signed

6 unsigned / -2 signed

7 unsigned / -1 signed

3+2 = -5 (too large for 3 bits)

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Condition Code Register

Bit Name Meaning after arithmetic N negative result is negative

Z zero result is zero

V overflow signed overflow

C carry carry or unsigned overflow

Condition code (CC) register inherently contains information about previous arithmetic or logical operation. Specific bits in the CC register are set under specific conditions.

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Examples of 4 bit overflow

1100 0110 +

(1)0010 = 210

= 1210 = 610

= 210

= - 410 = 610

0100 0100 +

(0)1000 = 810

= 410 = 410

= - 810

= 410 = 410

1000 1000 +

(1)0000 = 010

= 810 = 810

= - 810 = - 810

= 010

0001 0001 +

(0)0010 = 210

= 110 = 110

= 110 = 110

= 210

Adapted from notes from BYU ECE124

32

Binary Multiplication

Multiplication Table 0 * 0 = 0 0 * 1 = 0 1 * 0 = 0 1 * 1 = 1

1210 = 610 =

1100 0110 * * 0000

1100 1100

0000 + 1001000 =7210

Adapted from notes from BYU ECE124

33

Introduction to moving data

l  CPU can load and store data into internal registers for faster operations

l  Done with first assembly instructions – mov.b and mov.w

mov.b #FFH, R5 mov.b R5, &P1OUT mov.w 2(R5), R6 mov.w @R5, 3(R6)

Adapted from notes from BYU ECE124

34

Logic review – basic gates

A Z 0 1 1 0

A B Z 0 0 0 0 1 0 1 0 0 1 1 1

A B Z 0 0 0 0 1 1 1 0 1 1 1 1

INV.W R5 R5 = ~R5;

BIS.B R6, R7 R7 = R6 || R7;

AND.W R6, R7 R7 = R6 && R7;

A B Z 0 0 0 0 1 1 1 0 1 1 1 0

XOR.W R6, R7 R7= R7 ^ R6;

A B Z 0 0 0 0 1 0 1 0 0 1 1 1

BIC.W R6, R7 R7 = !R6 && R7;

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Logic review – example

BIS.B #08h, R5

BIC.B #08h, R5

C and Assembly code to set bit 3 of local variable - var1 in R5

C and Assembly code to clear bit 3 of local variable - var1 in R5

Var1 = Var1 && $F7;

Var1 = Var1 || $08;

36

Shift Operations

c

c 0

c

c

RRA.B

RLA.B

RRC.B

RLC.B

divide by 2 – signed

multiply by 2 – signed or unsigned C command X = Y << 1;

Rotate right through C

Rotate left through C

37

Shift Operations

Multiplying by a power of 2 is shifting to the left. M • 2n will shift binary number M left by N bits M / 2n = M • 2-n will shift binary number M right by N bits

When dividing, be careful about the most significant bit. If signed, the existing bit must be replicated for sign extension. If unsigned, the most significant bit must always be a zero

38

Encoders / Decoders

3 to 8 line decode

input 0 1 2 3 4 5 6 7 000 1 0 0 0 0 0 0 0 001 0 1 0 0 0 0 0 0 010 0 0 1 0 0 0 0 0 011 0 0 0 1 0 0 0 0 100 0 0 0 0 1 0 0 0 101 0 0 0 0 0 1 0 0 110 0 0 0 0 0 0 1 0 111 0 0 0 0 0 0 0 1

Encoder Decoder

39

Read Only Memories - ROM

Address Data Out 1 2 2 3 4 5 6 7

000 1 0 1 0 1 0 0 0 001 0 1 0 0 0 0 0 1 010 0 0 1 0 0 1 0 0 011 0 0 1 1 0 0 0 0 100 0 0 1 0 0 0 1 1 101 1 1 1 0 0 1 1 1 110 1 1 0 0 0 0 1 0 111 1 1 1 1 0 0 0 1

n

m

ROM

2n addresses addressed by n bit m output data bits

40

Read Only Memories - ROM

n

m

Memory array n to 2n decoder

2n

Address input word lines

data output

41

ROM memory array

pd pd pd pd pd pd pd pd pd pd word line 2N-1

word line 2N-2

word line 2

word line 1

word line 0

data m-1 data m-2 data0 data1 data 2

programmed “one”

programmed “zero”

bit lines

weak pull down

42

Random Access Memory (RAM)

wordline

bitline bitlineN

6 Transistor SRAM cell

n

m

sram array of 6-t cells n to 2n decoder

2n Address input

word lines m

rnw

data in

data out 43

Creating Fractions

l  Fractions are created by using extra bits below your whole numbers.

l  The programmer is responsible for knowing where the “decimal place” is.

l  Move the decimal place by using the shift operator (<< or >>). –  Shifting is multiplying by powers of 2. –  Example:

l  x << 5 = x × 25 l  x >> 5 = x × 2-5

44

Fractional Example

0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 = 5

A/D Sample (10-bit)

Fractional part

Shift left by 6 (i.e. A2D << 6;):

0

Whole part

0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 = "5.0"

45

Fractional Example l  We know 5/2 = 2.5

–  If we used pure integers, 5/2 = 2 –  Using a fixed-point fractional portion can recover the

lost decimal portion.

0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 = 5/2 = 2

A/D Sample (10-bit)

Fractional part

Whole part

0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 = 5.0/2 = 2.5 1

46

Fixed Point

Fractional part

Whole part

0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1

§  By using a fixed-point fractional part, we can have 5/2 = 2.5

§  The more bits you use in your fractional part, the more accuracy you will have.

§  Accuracy is 2^-(fraction bits).

§  For example, if we have 6 bits in our fractional part (like the above example), our accuracy is 2^-6 (2-6 = 0.015625. In other words, every bit is equal to 0.015625

47

Fixed Point Arithmetic

Adding 2.5 + 2.5 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1

§  Fixed point addition:

0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1

0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0

+

0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 = 5 0

§  Shift right to regain original resolution and data position.

§  Without fixed point math the result would have been 4 due to the truncation of the integer division. 48

Fixed Point Example #1

Cruise Control Specification range 0 – 100 MPH, resolution of 1 MPH two values: measured speed and target speed algorithm constantly calculates difference Format determination Use binary fixed point for ease of math (subtraction) Select range of 0 to 127 (27) to just exceed requirements Select resolution of 0.5 (2-1) to just exceed requirements 7 bits for integer and 1 bit for fractional = 8 bits precision

0 1 0 0 1 1 0 0

implied point

integer part fractional part

= 70.0 = K • Δp = 140 • 2-1

49

Fixed Point Example #2

A/D converter Specification range 0 – 5 V, resolution of 20 millivolts Format determination Select range of 0 to 7 (23) to just exceed requirements Select resolution of 0.015625 (2-6) to just exceed rqmts 3 bits for integer and 6 bit for fractional = 9 bits precision for 9 bits precision, need 16 bit register in HCS12

0 1 1 1 1 0 0 0

implied point

integer part fractional part

= 3.75 = K • Δp = 240 • 2-6 0

50

Floating Point Numbers

l  Binary scientific notation l  32-bit floating point

l  Exponent is biased l  Implied leading 1 in mantissa

s exponent mantissa 1 8 23

1272.11 −××−= exponents fractionN

51

Floating Point Numbers

l  What does this represent?

Positive number

Exponent is 128 which means the real exponent is 1

Mantissa is to be interpreted as 1.1 This is 20 + 2-1 = 1 + 1/2 = 1.5

The final number is 1.5 x 21 = 3

0 10000000 10000000000000000000000

52

ASCII Characters

NUL DLE SP 0 @ P ` p SOH DC1 ! 1 A Q a q STX DC2 “ 2 B R b r ETX DC3 # 3 C S c s EOT DC4 $ 4 D T d t ENQ NAK % 5 E U e u ACK SYN & 6 F V f v BEL ETB ‘ 7 G W g w BS CAN ( 8 H X h x HT EM ) 9 I Y i y LF SUB * : J Z j z VT ESC + ; K [ k { FF FS , < L \ l | CR GS - = M ] m } SO RS . > N ^ n ~ SI US / ? O _ o DEL

0 1 2 3 4 5 6 7 8 9 a b c d e f

0 1 2 3 4 5 6 7 8-9 a-f

53

Properties of ASCII Code l  What is relationship between a decimal digit ('0', '1', …)

and its ASCII code? l  What is the difference between an upper-case letter

('A', 'B', …) and its lower-case equivalent ('a', 'b', …)? l  Given two ASCII characters, how do we tell which

comes first in alphabetical order? l  What is significant about the first 32 ASCII codes? l  Are 128 characters enough? (http://www.unicode.org/)

54

Data Representation – endian

l  Byte Ordering –  Little endian

l  refers to system where bytes start in memory from small to large

l  Intel and HP –  Big endian

l  refers to system where bytes start in memory from large to small

l  IBM and Motorola (this class) –  Biendian - goes either way – PowerPC

l  Bit Ordering $00

$01

$02

$03

$12

$34

$56

$78

$12345678 $00

$01

$02

$03

$78

$56

$34

$12 big endian little endian

55

Data representation – BCD

1 2 : 4 5 0001 0010 : 0100 0101

l  Binary Coded Decimal –  useful for human readable displays (7 segment) –  half carry bit for BCD arithmetic –  values 1010 through 1111 are not legal –  H CCR bit and DAA instructions used with BCD

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